91影视

In linear algebra, a symmetric matrix is a square matrix that stays unchanged when its transpose is calculated. In that instance, a symmetric matrix is one whose transpose equals the matrix itself.

We have to show that for every symmetric $n脳n$matrix A , there exists a symmetric $n脳n$matrix B such that $B^3=A$. Now consider any symmetry $n脳n$matrix A . Therefore, there exists an orthogonal matrix and a diagonal matrix such that:

$A=SD{S}^{-1}$

Now, consider the cube root of the diagonal matrix D,

$D=D_0^3$(for some diagonal matrix$D_0$)

Now consider role="math" localid="1659616689432" $$\begin{gathered} A=SD{S}^{-1} \\ 鈬�A=S\left(D_0^3\right)S^{-1}(SubstituteD=D_0^3) \\ 鈬�A={\left(S{D}_0{S}^{-1}\right)}^3 \\ 鈬�A=B^3(whereB=S{D}_0{S}^{-1}) \end{gathered}$$as follows:

Since $B=S{D}_0{S}^{-1},B$is a symmetric $n脳n$matrix. Therefore, we can conclude that for every symmetric $n脳n$matrix A , there exists a symmetric $n脳n$matrix B that $B^3=A$.